chapter 7 - centroids & centers of mass

Chapter 7 - Centroids & Centers of MassEngineering Mechanics: STATICS Fifth Edition in SI Units
Chapter 7: Centroids & Centers of Mass
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Copyright 2008 Pearson Education South Asia Pte Ltd
Learning Objective
To be balanced, the woman’s center of mass – the point at which her weight effectively acts – must be directly above her hands.
This chapter introduces the concept of an average position, or centroid, and shows how to locate the centers of mass of objects.
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Chapter Outline
Center of Mass of Composite Objects
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Centroids of Areas
To determine the average position of a group of students sitting in a room:
Introduce a coordinate system to specify the position of each student
E.g. align the axes with the walls of the room
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Centroids of Areas
Number the students from 1 to N & denote the position of student 1 by (x1, y1), the position of student 2 by (x2, y2) & so on
The average x coordinate, which is denoted by , is the sum of their x coordinates divided by N:
(7.1)
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Centroids of Areas
(7.2)
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Centroids of Areas
Suppose that we pass out some pennies to the students:
Let the number of coins given to student 1 be c1, the number given to student 2 be c2 & so on
The average position of the coins may not be the same as the average position of the students
E.g. if the students in the front of the classroom have more coins, the average position of the coins will be closer to the front of the front than the average position of the students
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Centroids of Areas
To determine the x coordinate of the average position of the coins, we need to sum the x coordinates of the coins by multiplying the number of coins each student has by his or her x coordinate & summing
We can obtain the number of the coins by summing the numbers c1, c2,…
Thus, the average x coordinate of the coins is:
(7.3)
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Centroids of Areas
Determine the average y coordinate of the coins in the same way:
(7.4)
By assigning other meanings to c1, c2,…, we determine the average positions of other measures associated with the students
E.g. we could determine the average position of their age or the average position of their height
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Centroids of Areas
More generally, we can use Eqns (7.3) & (7.4) to determine the average position of any set of quantities with which we can associate positions
An average position obtained from these equations is called a weighted average position or centroid
The “weight” associated with position (x1, y1) is c1, the “weight” associated with position (x2, y2) is c2 & so on
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Centroids of Areas
Consider an arbitrary area A in the x-y plane:
Divide the area into parts A1, A2,…, AN & denote the positions of the parts by (x1, y1), (x2, y2),…, (xN, yN)
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Centroids of Areas
The centroid or average position of the area, by using Eqns (7.3) & (7.4) with the areas of the parts as the weights:
(7.5)
To reduce the uncertainty in the positions of areas A1, A2,…, AN, divide A into smaller parts:
But we would still obtain only approximate values of
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Centroids of Areas
To determine the exact location of the centroid, we must take the limit as the sizes of the parts approach zero:
We obtain this limit by replacing Eqns (7.5) by the integrals:
(7.6)
(7.7)
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Centroids of Areas
Where x & y are the coordinates of the differential element of area dA
The subscript A on the integral sign means the integration is carried out over the entire area
The centroid of the area is:
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Centroids of Areas
Keeping in mind that the centroid of an area is its average position, will often help you locate it:
E.g. the centroid of a circular area or a rectangular area obviously lies at the center of the area
If an area has “mirror image” symmetry about an axis, the centroid lies on the axis
If an area is symmetric about 2 axes, the centroid lies at their intersection
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Example 7.1
Centroid of an Area by Integration
Determine the x coordinate of the centroid of the triangular area.
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Example 7.1 (continued)
Strategy
Evaluate Eqn (7.6) using an element of area dA in the form of a vertical “strip” of width dx.
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Solution
The height of a strip of width dx at position x is (h/b)x, so its area is dA (h/b)x dx. Use this expression to evaluate Eqn (7.6).
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Practice Problem
Determine the y coordinate of the centroid of the triangular area. Evaluate Eqn (7.7) by using an element of area dA in the form of a vertical “strip” of width dx, and let y be the height of the midpoint of the strip.
Answer:
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Example 7.2
(refer to textbook)
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Composite Areas
Although centroids of areas can be determined by integration, the process becomes difficult and tedious for complicated areas.
In this section describe a much easier approach that can be used if an area consists of a combination of simple areas, which we call a composite area.
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Composite Areas
Composite area: an area consisting of a combination of simple areas
The centroid of a composite area can be determined without integration if the centroids of its parts are known
The area in the figure consists of a triangle, a rectangle & a semicircle, which we call parts 1, 2 & 3
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Composite Areas
The x coordinate of the centroid of the composite area is:
(7.8)
From the equation for the x coordinate of the centroid of part 1:
We obtain:
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Composite Areas
Using this equation & equivalent equations for parts 2 & 3, we can write Eqn (7.8) as:
The coordinates of the centroid of a composite area with an arbitrary number of parts are:
(7.9)
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Composite Areas
The area in the figure consists of a triangular area with a circular hole or cutout:
Designate the triangular area (without the cutout) as part 1 of the composite area & the area of the cutout as part 2
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Composite Areas
The x coordinate of the centroid of the composite area is:
Therefore, we can use Eqn (7.9) to determine the centroids of composite areas containing cutouts by treating the cutouts as negative areas
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Composite Areas
Determining the centroid of a composite area requires 3 steps:
1. Choose the parts — try to divide the composite area into parts whose centroids you know or can easily determine.
2. Determine the values for the parts — determine the centroid & the area of each part. Watch for instances of symmetry that can simplify your task.
3. Calculate the centroid — use Eqn (7.9) to determine the centroid of the composite area.
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Example 7.3
Centroid of a Composite Area
Determine the x coordinate of the centroid of the composite area.
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Strategy
Divide the area into parts (the parts are obvious in this example), determine the centroids of the parts & apply Eqn (7.9).
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Choose the Parts:
Divide the area into simple parts. The x coordinates of the centroids of the parts are shown.
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Tabulate the terms needed to apply Eqn (7.9).
Refer to textbook Appendix B.
Ai
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Calculate the Centroid
Use Eqn (7.9) to determine the x component of the centroid.
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Practice Problem
Determine the y coordinate of the centroid of the composite area.
Answer:
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Example 7.4
(refer to textbook)
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Distributed Loads
In many engineering applications, loads are continuously distributed along lines:
E.g. The load exerted on a beam supporting a floor of a building is distributed over the beam’s length
The load exerted by wind on a television transmission tower is distributed along the tower’s height
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Distributed Loads
The concept of the centroid of an area is useful in the analysis of objects subjected to such loads
Describing a Distributed Load:
Suppose that we pile bags of sand on a beam:
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Distributed Loads
It is clear that the load exerted by the bags is distributed over the length of the beam & that its magnitude at a given position x depends on how high the bags are piled at that position
To describe the load, we define a function w such that the downward force exerted on an infinitesimal element dx of the beam is w dx
With this function we can model the varying magnitude of the load exerted by the sand bags:
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Distributed Loads
The arrows in the figure indicate that the load acts in a downward direction
Loads distributed along lines, from simple examples such as the beam’s own weight to complicated ones such as the lift distributed along the length of an airplane’s wing, are modeled by the function w
Since the product of w & dx is a force, the dimensions of w are (force)/(length)
w can be expressed in newtons per meter (SI units)
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Distributed Loads
Determining Force & Moment:
Assuming that the function w describing a particular distributed load is known as:
The graph of w is called the loading curve
The force acting on an element dx of the line is w dx
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Distributed Loads
The total force F exerted by the distributed load is determined by integrating the loading curve with respect to x:
(7.10)
We can also integrate to determine the moment about a point exerted by the distributed load
The moment about the origin due to the force exerted on the element dx is xw dx
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Distributed Loads
The total moment about the origin due to the distributed load:
(7.11)
When you are concerned only with the total force & moment exerted by a distributed load, you can represent it by a single equivalent force F:
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Distributed Loads
For equivalence, the force must act at a position on the x axis such that the moment of F about the origin is equal to the moment of the distributed load about the origin:
Therefore the force F is equivalent to the distributed load if we place it at the position:
(7.12)
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Distributed Loads
The Area Analogy:
Notice that the term w dx is equal to an element of “area” dA between the loading curve & the x axis (w dx is actually a force & not an area)
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Distributed Loads
Interpreted in this way, Eqn (7.10) states that the total force exerted by the distributed load is equal to the “area” A between the loading curve & the x axis:
(7.13)
(7.14)
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Distributed Loads
The force F is equivalent to the distributed load if it acts at the centroid of the “area” between the loading & the x axis
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Example 7.5
Beam with a Distributed Load
The beam is subjected to a “triangular” distributed load whose value at B is 100 N/m. (That is, the function w increases linearly from w = 0 at A to w = 100 N/m at B.) Determine the reactions on the beam at A and B.
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Strategy
Use the area analogy to represent the distributed load by an equivalent force. Then we can apply the equilibrium equations to determine the reactions at A and B.
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Solution
The “area” of the triangular distributed load is one-half its base times its height or
The centroid of the triangular “area” is located at:
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Solution
We can therefore represent the distributed load by an equivalent downward force of 600-N magnitude acting at x = 8 m:
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Σ Mpoint A = (12 m) B (8 m)(600 N) = 0
we obtain Ax = 0, Ay = 200 N & B = 400 N
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Practice Problem
Determine w as a function of x for the triangular distributed load in this example.
Use Eqns (7.10) and (7.11) to determine the total downward force and the total clockwise moment about the left end of the beam due to the triangular distributed load.
Answer:
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Example 7.6
(refer to textbook)
(refer to textbook)
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Centroids of Volumes and Lines
Definitions:
Volumes:
Consider a volume V & let dV be a differential element of V with coordinates x, y & z:
By analogy with Eqns (7.6) & (7.7), the coordinates of the centroid V are:
(7.15)
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The subscript V on the integral sign means that the integration is carried out over the entire volume
If a volume has the form of a plate with uniform thickness & cross-sectional area A, its centroid coincides with the centroid of A & lies at the midpoint between the 2 faces
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To show that this is true, we obtain a volume element dV by projecting an element dA of the cross-sectional area through the thickness T of the volume, so that dV = T dA
The x & y coordinates of the centroid of the volume are:
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The coordinate by symmetry
Thus you know the centroid of this type of volume if you know (or can determine) the centroid of its cross-sectional area
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Lines:
The coordinates of the centroid of a line L are:
(7.16)
length of the line with the
coordinates x, y & z.
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Example 7.8
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Strategy
Because of the axial symmetry of the cone, the centroid must lie on the x axis. We will determine the x coordinate of the centroid by applying Eqn (7.15), using an element of volume dV in the form of a disk of thickness dx.
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An element of volume in the form of a disk.
The radius of the disk at position x is (R/h)x. The volume of the disk is the product of the area of the disk and its thickness:
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Practice Problem
The radius in metres of the circular cross section of the
truncated cone is given as a function of x by .
Determine the x coordinate of its centroid.
Answer: 2.43 m
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Example 7.9
(refer to textbook)
(refer to textbook)
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Refer To Textbook Chapter 7
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chapter 7 - centroids & centers of mass

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